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SNIIX18
 2 years ago
If the period of a pendulum is 2.0 seconds and the acceleration due to gravity is 9.8 meters/second^2, What is the length of the cord on the pendulum?
A)1.0 meters
B)0.67 meters
C)1.3 meters
D)3.5 meters
SNIIX18
 2 years ago
If the period of a pendulum is 2.0 seconds and the acceleration due to gravity is 9.8 meters/second^2, What is the length of the cord on the pendulum? A)1.0 meters B)0.67 meters C)1.3 meters D)3.5 meters

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theEric
 2 years ago
Best ResponseYou've already chosen the best response.1Are you speaking of a simple pendulum?

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1Then you'll want the equation that uses math to describe the relationships between different physical quantities. I think I've show it to you before. I found it at http://hyperphysics.phyastr.gsu.edu/hbase/pend.html It is:\[T=2\pi\sqrt{\frac{L}{g}}\] \(T\) is the period. \(L\) is the length of the chord. \(g\) is the acceleration caused by gravity. You want to solve for \(L\).

SNIIX18
 2 years ago
Best ResponseYou've already chosen the best response.0I GOT O.67 IS THAT RIGHT

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1It's not what I got! How did you solve for \(L\)?

SNIIX18
 2 years ago
Best ResponseYou've already chosen the best response.0I squared 2 then i divided 2.0 by 9.8

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1I'm confused... Can you show me what you got? You squared the 2?

SNIIX18
 2 years ago
Best ResponseYou've already chosen the best response.0now im confused cause your confused

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1Haha, you start with \[T=2\pi\sqrt{\frac{L}{g}}\]Now, what did you square?

SNIIX18
 2 years ago
Best ResponseYou've already chosen the best response.0owwww okay i see what i did wrong i was suppost to pi the 2 instead of square

theEric
 2 years ago
Best ResponseYou've already chosen the best response.1I would divide the equation by \(2\pi\), square the equation, and multiply the equation by \(g\), if you know what I mean. Then you have \(L\) by itself after applicable cancelations, if you know what I mean.
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